R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: EXERCISE 11

In a parallelogram $ABCD$, any point $E$is taken on the side $BC$. $AE$ and $DC$when produced meet at a point $M.$Prove that
$ar(∆ADM)=ar\left(ABMC\right)$

In a triangle$ABC$, the medians$BE$and $CF$ intersect at $G.$ Prove that $ar(∆BCG)=ar\left(AFGE\right).$

In the adjoining figure, $BD\parallel CA$, $E$ is the midpoint of$CA$and$BD=\frac{1}{2}CA$ . Prove that $ar(∆ABC)= 2ar(∆DBC).$

In the adjoining figure,$CE \left|\right| AD$ and$CF \left|\right| BA$. Prove that $ar(∆CBG)=ar(∆AFG).$

In a trapezium $ABCD$,$AB \left|\right| DC, AB=a \mathrm{cm},$ and $DC=b \mathrm{cm}$. If $M$and $N$ are the midpoints of the nonparallel sides, $AD$ and $BC$ respectively then find the ratio of $ar\left(DCNM\right) \mathrm{and} ar\left(MNBA\right).$

$ABCD$ is a trapezium in which $AB \left|\right| CD, AB=16 \mathrm{cm} \mathrm{and} DC=24 \mathrm{cm}$. If $E$ and $F$ are respectively the midpoints of $AD$and $BC$, prove that $ar\left(ABFE\right) =\frac{ 9}{11} ar\left(EFCD\right).$

In the adjoining figure, $D$ and $E$ are respectively the midpoints of sides$AB$ and $AC$of $∆ABC$. If $PQ \left|\right| BC$and $CDP \mathrm{and} BEQ$are straight lines then prove that $ar(∆ABQ)=ar(∆ACP).$

In the adjoining figure,$ABCD$ and$BQSC$ are two parallelograms. Prove that $ar(∆RSC)=ar(∆PQB).$